第6回平均の差の検定

Data Analysis and Experimental Design
第６回平均の差の検定
H0：μ= a
T=
x −a
s2
n
> T <- (100.5-100)/(sqrt(1.0/50));T
[1] 3.535534
> qt(0.975, 49)
[1] 2.009575
Data Analysis and Experimental Design
H0：μ1= μ2 T =
x1 − x2
(n1 − 1)s12 + (n2 − 1)s22 ⎛⎜ 1
n1 + n2 − 2
1⎞
⎜ n + n ⎟⎟
2 ⎠
⎝ 1
df = n1 + n2 - 2
1
Data Analysis and Experimental Design

n1 = n2 = nの時は次式となる（df = 2n - 2）：
x1 − x2
T=
(s12 + s22 )
n
> T <- (72.0-76.0)/(sqrt((36.0+39.0)/8));T
[1] -1.306395
> qt(0.025, 14)
[1] -2.144787
Data Analysis and Experimental Design
T=
x1 − x2
s12 + s22 − 2rs1s2
n
n
T=
∑D
i =1
i
⎛ n
⎞
n∑ D − ⎜ ∑ Di ⎟
i =1
⎝ i =1 ⎠
n −1
n
2
2
i
Di = x1i – x2i
2
Data Analysis and Experimental Design
> T <- (72-76)/(sqrt((41.1+44.6+ 2*0.894*sqrt(41.1)*sqrt(44.6))/8))
> T
[1] -3.740581
> qt(0.025, 7)
[1] -2.364624
3

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